Remember:
1
2
L± |l, ml i = (l ± ml + 1)(l ⌥ ml ) ~|l, ml ± 1i
~J 2 = ~L2 + ~S 2 + 2Lz Sz + L+ S + L S+
We denote the eigenstates of ~J 2 , Jz , ~L2 , ~S 2
with |j, m, l, 12 i thus |j, m, li
Proposition:
1
1
|l + , l + , li = |l, li|"i
2
2
Proof:
1
Jz |l, li|"i = (Lz + Sz )|l, li|"i = ~(l + )|l, li|"i X
2
~J 2 |l, li|"i = ~2 l(l + 1) + 1 ( 1 + 1) + 2l 1 |l, li|"i
2 2
2
1
3
2
= ~ (l + )(l + )|l, li|"i X
2
2
Now consider J |l, li|"i
1
2
J |l, li|"i =(2l) ~|l, l
1
2
2
Jz J |l, li|"i =(2l) ~ (l
1
)|l, li|#i
2
1
)J |l, li|"i
2
=~(l
[~J 2 , J ] = 0
1i|"i + ~|l, li|#i
1
1) + |l, l 1i|"i + ~2 (l
2
) ~J 2
has eigenvalue
1
l+
2
normalize:
1
|l + , l
2
1
, li =
2
r
2l
|l, l
2l + 1
1i|"i +
mathematical induction:
s
l + mj + 12
1
|l + , mj , li =
|l, mj
2
2l + 1
mj : l + 12 , ..., (l + 12 ) half-integer values
r
1
i|"i +
2
1
|l, li|#i
2l + 1
s
l
mj + 12
1
|l, mj + i|#i
2l + 1
2
(⇤)
1
2 , ...,
construct states with mj : l
|l
1
, mj , li =
2
where mj : l
1
2 , ...,
s
(l
l
(l
mj + 12
|l, mj
2l + 1
1
2)
which are orthogonal to (*):
1
i|"i +
2
s
l + mj + 12
1
|l, mj + i|#i
2l + 1
2
1
2 ).
We have:
Jz |l
~J 2 |l
1
1
, mj , li =~mj |l
, mj , li (obvious)
2
2
1
1
1
1
2
, mj , li =~ (l
)(l + )|l
, mj , li
2
2
2
2
thus eigenstates of ~J 2 and Jz with j = l
1
2
(exercise)
and mj and of ~L2 and ~S 2 .
For l = 0 there is just | 21 , ± 12 , 0i
The state |0 12 , mj , 0i does not exist.
Summary:
1
1
1
|l ± , mj , li = ↵± |l, mj
i|"i + ± |l, mj + i|#i
2
2
2
s
l ± mj + 12
2
where ↵± = ±
= ± ⌥ and +
+ 2 =1
2l + 1
The general case
~J = ~J1 + ~J2
complete set of commuting angular momentum operators
~J 2 , J1z , ~J 2 , J2z
1
2
~J 2 , Jz , ~J 2 , ~J 2
1
2
corresponding eigenfunctions: orthonormal system
|j1 m1 i|j2 m2 i
|jmj1 j2 i
(⇤⇤)
(⇤)
expand (*) in terms of (**). Expansion coefficients:
hj10 m1 j20 m2 |jmj1 j2 i = hj10 m1 |hj20 m2 |jmj1 j2 i
consider:
hj10 m1 j20 m2 |(~Ji2 |jmj1 j2 i) = (hj10 m1 j20 m2 |~Ji2 )|jmj1 j2 i
) ji0 (ji0 + 1)hj10 m1 j20 m2 |jmj1 j2 i = ji (ji + 1)hj10 m1 j20 m2 |jmj1 j2 i
)
only coefficients with j10 = j1 and j20 = j2 are not equal zero.
Now consider matrix element of Jz = J1z + J2z
hj1 m1 j2 m2 |Jz |jmj1 j2 i =(m1 + m2 )hj1 m1 j2 m2 |jmj1 j2 i
=mhj1 m1 j2 m2 |jmj1 j2 i
)
)
only coefficients with m = m1 + m2 do not vanish.
expansion of |jmj1 j2 i in terms of {|j1 m1 j2 m2 i}
X
|j1 m1 j2 m2 i hj1 m1 j2 m2 |jmj1 j2 i
|jmj1 j2 i =
|
{z
}
m
1
m2 =m m1
Clebsh-Gordon coefficients
Determine j for given j1 and j2
m
j1 + j2
j1 + j2
j1 + j2
..
.
j1
j1
..
.
j2
j2
1
2
(m1 , m2 )
(j1 , j2 )
(j1 1, j2 ), (j1 , j2 1)
(j1 2, j2 ), (j1 1, j2 1), (j1 , j2
..
.
1
(j1
(j1
..
.
2j2 , j2 ), ..., (j1 , j2 2j2 )
2j2 1, j2 ), ..., (j1 1, j2 )
2)
degeneracy of m
1
2
3
..
.
2j2 + 1
2j2 + 1
..
.
Degeneracy of m
m
m |j1 j2 |
|j1 j2 | < m < |j1
m  |j1 j2 |
j2 |
Degree
j1 + j2
j1 + j2
j1 + j2
of degeneracy
m+1
|j1 j2 | + 1
|m| + 1
Multiplet structure
value of j
j1 + j2
j1 + j2 1
..
.
corresponding m-value
j1 + j2 , ..., (j1 + j2 )
j1 + j2 1, ..., (j1 + j2
..
.
|j1
|j1
j2 |
j2 |, ..., |j1
possible values of j: |j1
1)
j2 |
j2 |  j  j1 + j2
Number of states is suitable: (j1
jX
1 +j2
(2j + 1) =
2j2
X
2(j1
j2 )
j2 + k) + 1
k=0
j=|j1 j2 |
= (2(j1 + j2 ) + 1)(2j2 + 1)
= (2j1 + 1)(2j2 + 1)
2j2 (2j2 + 1)
which clearly agrees with the number of states (|j1 m1 i|j2 m2 i) X
Wigner-3j-symbol
◆
✓
j1
j2
j3
= ( 1)j1
m1 m2 m3
Clebsch-Gordon for ~L
j2 m3
hj1 m1 j2 m2 |j3 m3 j1 j2
p
2j3 + 1
~S:
j1 = l, j2 = s = 12 , j = l + 12 , l
1
2
hjml 12 |lm1 12 m2 i
j
l+
l
1
2
1
2
m2 =
✓
1
2
l+m+1/2
2l+1
✓
◆1/2
l m+1/2
2l+1
◆1/2
m2 =
✓
✓
1
2
l m+1/2
2l+1
l+m+1/2
2l+1
◆1/2
◆1/2
Approximation Methods for Stationary States
perturbation theory:
problem di↵ers only slightly from an exactly solvable problem
variatonal method:
calculation of ground state energy
WKB-method:
nearly classical limit
Time Independent Perturbation Theory (Rayleigh-Schrödinger)
Hamilton operator consists of two parts:
H = H0 + H1
Eigenvalues En0 and eigenfunctions |n0 i of H0 are known
H0 |n0 i = En0 |n0 i
One seeks: En , |ni:
H|ni = En |ni
(⇤)
Power series expansion in the parameter
En =En0 + En1 +
2
|ni =|n0 i + |n1 i +
En2 + ...
2
|n2 i + ...
(not possible in general):
(⇤⇤)
Nondegenerate Perturbation Theory
From (⇤), (⇤⇤) we get:
(H0 + H1 )(|n0 i + |n1 i +
2
|n2 i + ...)
= (En0 + En1 + En2 + ...)(|n0 i + |n1 i +
Comparison of the coefficients (using (*)):
0
H0 |n i =
H0 |n1 i + H1 |n0 i =
H0 |n2 i + H1 |n1 i =
..
.
En0 |n0 i
En0 |n1 i
En0 |n2 i
not known
+ En1 |n0 i
+ En1 |n1 i + En2 |n0 i
not known
Normalization of |ni (convenient):
hn0 |ni = 1 )
)
hn0 |n1 i +
2
hn0 |n2 i + ... = 0
hn0 |n1 i = hn0 |n1 i = ... = 0
2
|n2 i + ...)
multiply (⇤) by hn0 |:
hn0 |H0 |n1 i +hn0 |H1 |n0 i = 0 + En1
| {z }
=0
)
En1 = hn0 |H1 |n0 i
unperturbed states |m0 i form a complete orthonormal set:
X
1
0
0 1
)
|n i =
cm |m i , cm = hm |n i
m6=n
multiply (⇤) by hm0 |:
hm0 |H0 |n1 i + hm0 |H1 |n0 i =En0 hm0 |n1 i + En1 hm0 |n0 i
(Em0
En0 ) hm0 |n1 i =
| {z }
cm
hm0 |H1 |n0 i + En1 hm0 |n0 i
now m0 6= n0 :
cm (En0
)
Em0 ) = hm0 |H1 |n0 i
X hm0 |H1 |n0 i
0
|n1 i =
|m
i
0
0
En Em
m6=n
energy in second order: multiplying by hn0 |:
hn0 |H0 |n2 i + hn0 |H1 |n1 i = En0 hn0 |n2 i + En1 hn0 |n1 i + En2 hn0 |n0 i
thus:
X |hm0 |H1 |n0 i|2
En2 = hn0 |H1 |n1 i =
En0 Em0
m6=n
Perturbation Theory for Degenerate States
Let |na0 i, |nb0 i, ... |nk0 i be degenerate:
0
H0 |ni i
=
0
✏|ni i
Search for new basis
0
0
hn↵ |H1 |n i
=
0
|n↵ i
with
(↵)
H1 ↵
then no sinqularities occur
X
Matrix elements in old basis
H1ij =
0
0
hni |H1 |nj i
(hermitian matrix)
The new states
X
0
|n↵ i =
ci↵ |ni0 i
i
give the matrix elements
0
H1↵ = hn↵
|H1 |n0 i =
=
X
i,j
X
⇤
hni0 |ci↵
H1 cj |nj0 i
⇤
ci↵
H1ij cj
i,j
thus
X
(↵)
⇤
ci↵
H1ij cj = H1↵ = H1
↵
i,j
with ci↵ representing an unitary transformation:
X
⇤
ci = ↵
= ci↵
i
X
↵
⇤
= ci↵ cj↵
=
ij
ki
X
ck↵
↵
X
X
⇤
ci↵
H1ij cj
=
X
(↵)
ck↵ H1 ↵
↵
i,j
H1kj cj =
( )
c k H1
(⇤)
j
For each
this is an eigenvalue equation.
solvability condition:
det(H1ij
(⇤)
)
( )
H1 ij )
ci↵
)
=0
)
0
|n↵
i
afterwards the same way as before.
( )
H1
Variational Principle (Ritz)
Hamilton operator H, basis |ni, arbitrary | i:
X
X
h |H| i =
h |nihn|H| i =
En h |nihn| i
n
n
=
X
n
En |h |ni|
=E0 h | i
)
h |H| i
E0 
h | i
2
E0
X
n
|h |ni|
2
choose an ansatz function | (µ)i depending on parameters µ
and minimize the expression
h (µ)|H| (µ)i
E (µ) =
h (µ)| (µ)i
upper bound for ground state energy
quadratic error in the energy, because, let
| i = |ni + |✏i
)
,
hn|✏i = 0
h |H| i
En + h✏|H|✏i
=
= En + O(✏2 )
h | i
hn|ni + h✏|✏i
important tool in mathematical physics in the proof of inequalities
The WKB (Wentzel-Kramers-Brillouin) Method
Consider stationary states of a potential.
high energies, so that typical wavelengths of the state are small in comparison to a
typical length of the potential.
Ansatz:
(~x ) = A(~x )e iS(~x )/~
substituting into Schrödinger equation
~2 2
r
2m
)
= (E
A(rS)2
V (~x ))
i~Ar2 S
2i~rA rS
~2 r2 A = 2m(E
V )A
real and imaginary parts:
(rS)2 =2m(E
V ) + ~(r2 A)/A
r2 S =2rS · r log A
Approximation: neglect ~2 (r2 A)/A
Now: 1D (includes radial movement)
from (⇤):
⇣
⌘
d 1
dS
log
+ log A = 0
dx 2
dx
C
)A= p
S0
(⇤⇤)
(⇤)
from (⇤⇤):
dS
dx
)
)
!2
=2m(E
S(x) = ±
(x) =
Z
X
+,
Momentum: p(x) =
p
V (x))
x
dx
0
p
C±
p
p(x)
V (x 0 ))
Z x
n
o
exp ± i
dx 0 p(x 0 )/~
2m(E
2m(E
v (x))
E < V : (tunneling): solution decreases or grows exponentially
Z x
0
X C±
p
(x) =
exp ⌥
dx 0 (x 0 )/~
(x)
+,
where
p
(x) = 2m(V (x)
E)